**Separation Of Variables**

Are you looking for a solution for Solve The Given Differential Equation By Separation Of Variables. Dp Dt = P − P2. Then you remain at the right place, as discussed in this article. We thoroughly check each answer to a question to provide you with the correct answers.

The basic concept behind this question is the knowledge of derivatives, integration, and the rules such as the product and quotient integration rules. In mathematics, the separation of variables is any of several methods for solving ordinary and partial differential equations, in which algebra allows one to rewrite an equation so that each of two variables occurs on a different side of the equation.

**Also read:** workforce software monday

## Answer For The Solve The Given Differential Equation By Separation Of Variables. Dp Dt = P − P2

The question wants us to find a specific solution to the differential equation d𝑃 by d𝑡 is equal to the square root of 𝑃 times 𝑡 using the initial condition that 𝑃 of one is similar to two. We can see that our differential equation is almost already the product of a function in 𝑃 and the product of a process in 𝑡.

In fact, by using our laws of exponents, we can see this more clearly. d𝑃 by d𝑡 is equal to the square root of 𝑃 multiplied by the square root of 𝑡. Since this is a first-order differential equation that can remain written as the products of a function in 𝑃 and a function in 𝑡, we call this a separable differential equation.

To solve this, we want to separate our variables 𝑃 and 𝑡 onto opposite sides of the equation. We’ll start by dividing through by the square root of 𝑃. This gives us that one divided by the square root of 𝑃 times d𝑃 by d𝑡 is equal to the square root of 𝑡. Although d𝑃 by d𝑡 is not a fraction, when solving separable differential equations, we can treat it a little bit like a fraction.

**Also read:** error 1045 (28000): access denied for user ‘root’@’localhost’ (using password: yes)

### Using this

we get the equivalent statement one divided by the square root of 𝑃 d𝑃 is equal to the square root of 𝑡 d𝑡. We then want to integrate both sides of this equation. If We could then evaluate both integrals by using the power rule for integration.

This tells us if 𝑛 is not equal to the negative one, the integral of 𝑥 to the 𝑛th power concerning 𝑥 is similar to 𝑥 to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝑐. We add one to the exponent and then divide by this new exponent.

### To See Why We Can Do This

we know that one divided by the square root of 𝑃 is the same as saying 𝑃 to the power of negative one-half, and the square root of 𝑡 is the same as saying 𝑡 to the power of one-half. To integrate 𝑃 to the power of negative a half concerning 𝑃, we add one to our exponent and divide by this new exponent and add a constant of integration we will call 𝑐 one.

And we can simplify this since negative a half plus one equals one-half. And we can simplify this further since dividing by a fraction is the same as multiplying by the reciprocal. So, dividing by one-half is the same as multiplying by two.

This gives us two 𝑃 to the power of a half plus 𝑐 one. We then do the same to integrate 𝑡 to the power of a half concerning 𝑡. We add one to our exponent and then divide by this new exponent. And we add a constant of integration we will call 𝑐 two. If We can simplify this expression further. We can combine the constants 𝑐 one and 𝑐 two into a new constant we will call 𝑐. We can simplify this further since one-half plus one equals three over two.

**Also read:** error in match.names(clabs, names(xi)) : names do not match previous names

### Finally

As we did before, instead of dividing by three over two, we can multiply by two and divide by three. So, we now have the equation two 𝑃 to the power of half is equal to two-thirds 𝑡 to the control of three over two-plus 𝑐. And we can find the value of 𝑐 since the question gives us the initial condition 𝑃 of one equals two. 𝑃 is a function of 𝑡, so this tells us when 𝑡 is equal to one, 𝑃 is equivalent to two.

## Separation of Variables

Consider a differential equation that can remain written in the form

M[x]+ N[y] dy \dx = 0

where is a continuous function of alone and is a constant function of alone. As you saw in Section 6.2, for this type of equation, all terms can remain collected, and all terms with a solution can remain obtained by integration.

Such equations are said to be separable, and the solution procedure is called the separation of variables. Below are some examples of differential equations that are separable. Original Differential Equation Rewritten with Variables Separated.

**Also read: **This application has no explicit mapping for /error, so you are seeing this as a fallback.